What is the identity of the element with an electron configuration of 1s²2s²2p6 3s²3p²? How many
electrons are located in its highest main energy level and what orbitals are they located in?

pCO2 is an important factor that affects various aspects of water chemistry and the impacts of ocean acidification. When pCO2 increases, the concentration of total CO2 dissolved in water also increases. This leads to changes in pH, which decreases due to increasing atmospheric CO2 concentration.

When pCO2 rises, the concentration of only CO2 dissolved in water increases. The dissolved CO2 forms carbonic acid, which contributes to the acidification of the ocean. This increase in CO2 affects the equilibrium between CO2, HCO3-, and CO3^2-, shifting it towards higher levels of dissolved CO2 and H+ ions, resulting in a lower pH.

In terms of the impacts of ocean acidification on different organisms, the effects can vary depending on their specific needs. An organism that requires CO2 is likely to fare better under ocean acidification since the increase in dissolved CO2 can provide them with a favorable environment. However, organisms that rely on HCO3- or CO3^2- may fare worse under ocean acidification, as the lower pH interferes with the availability of these carbonate ions, which are essential for shell formation and calcification in some marine organisms.

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Answer:

3.5x1024 3.5 x 10 24 carbon dioxide molecules weighs 256 g.

Answer:

hydrocarbon

Explanation:

these are called hydrocarbons

Answer:

2.66 · 10⁵

Explanation:

There are 1440 minutes in a day. Which means that there is 266400 minutes in 185 days. And in scientific notation it is: 2.66 · 10⁵.

The most stable contributor's structure is the genuine structure of the resonance hybrid. The delocalization of electrons is what stabilizes the resonance hybrid. The potential resonance forms are quickly interconverted by a resonance hybrid.

Equivalent resonance types equally contribute to the hybrid's overall structure. Consider the resonance hybrid structure of a carboxylate group as an example. Different resonance contributors do not always contribute equally to the hybrid structure until they are equivalent to one another in terms of stability, as is the case for the carboxylate group, which has equivalent contributions from A and B as shown in the given figure. One resonance structure will more closely resemble the “actual” (hybrid) structure than another if it is more stable (lower in energy) than the other.

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= 0.2+560.66JK-2eF

use this app : Photomath

It will help you :)

Answer:

2.82*10^23

Explanation:

please see attached for work!

The balanced chemical equation for the reaction between nitrogen monoxide and oxygen gas is:

2NO(g) + O2(g) → 2NO2(g)

LR: NO

Excess: O2

Grams of target product produced: 30.7 g NO2

The balanced chemical equation for the reaction between nitrogen monoxide and oxygen gas is:

2NO(g) + O2(g) → 2NO2(g)

To determine the theoretical yield of nitrogen dioxide gas, we need to first identify the limiting reactant (LR) and the excess reactant. To do this, we can calculate the amount of product that would be produced from each reactant, assuming that they completely react:

From 22.0 g of NO: (22.0 g NO) / (30.01 g/mol NO) x (2 mol NO2 / 2 mol NO) x (46.01 g/mol NO2) = 30.7 g NO2

From 22.0 g of O2: (22.0 g O2) / (32.00 g/mol O2) x (2 mol NO2 / 1 mol O2) x (46.01 g/mol NO2) = 64.4 g NO2

Since NO is the limiting reactant, we can expect that 30.7 g of NO2 would be the theoretical yield of the reaction.

LR: NO

Excess: O2

Grams of target product produced: 30.7 g NO2

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Answer: 77.47 g of \(Fe_2O_3\) will be produced from 100.0 g of Iron (III) hydroxide

Explanation:

To calculate the moles :

\(\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}\)    

\(\text{Moles of} Fe(OH)_3=\frac{100.0g}{106.9g/mol}=0.935moles\)

The balanced chemical equation is:

\(2Fe(OH)_3(s)\rightarrow Fe_2O_3(s)+3H_2O(g)\)  

According to stoichiometry :

2 moles of \(Fe(OH)_3\) produce = 1 mole of \(Fe_2O_3\)

Thus 0.935 moles of \(Fe(OH)_3\) will producee=\(\frac{1}{2}\times 0.935=0.468moles\)  of \(Fe_2O_3\)

Mass of \(Fe_2O_3=moles\times {\text {Molar mass}}=0.468moles\times 159.7g/mol=74.74g\)

Thus 77.47 g of \(Fe_2O_3\) will be produced from 100.0 g of Iron (III) hydroxide

Answer:

375 grams

Explanation:

The percent yield if 155 grams of calcium carbonate is treated with 250 grams of hydrochloric acid and 142 grams of calcium chloride is obtained is 82.55%

Percent yield is defined as the amount of product actually made compared with the maximum calculated yield. It is the ratio of actual yield to the theoretical yield.

Mass of calcium carbonate = 155 g

Mass of HCl = 250g

Mass of calcium chloride product = 142 g

Moles of calcium carbonate = 155 / 100 = 1.55 moles

Moles of HCl = 250 / 36.5 = 6.85 moles

Mass of calcium chloride = Moles x Molar mass

                                          = 1.55 mole x 111 g/ mole

                                          = 172 g /mole

Theoretical yield = 172 g

Actual yield = 142 g

Percent yield = Actual yield / Theoretical yield x 100
                      = 142 / 172 x 100

                      = 82.55 %

Thus, the percent yield if 155 grams of calcium carbonate is treated with 250 grams of hydrochloric acid and 142 grams of calcium chloride is obtained is 82.55%

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Answer:

0.3kg/min

Explanation:

The formula for acceleration is a = f/m where f = force and m = mass.

a = 3/10

a = 0.3

Best of Luck

The value of electric flux is 2.45N.m²/C.

A flat sheet of paper of area 0.130 m² is oriented so that the normal to the sheet is at an angle of 56 ∘ to a uniform electric field of magnitude 18 N/C.

The electric flux through the paper is given by;

ϕ=E.A.cosθϕ

= Electric fluxE

= Electric field intensityA

= Area of the surfaceθ

= Angle between electric field intensity and normal to the surface

Given,E = 18 N/C A = 0.130 m² θ = 56°

The electric flux through the paper is :

ϕ=E.A.cosθϕ

= (18 N/C)(0.130 m²)cos56°ϕ

= 2.45 N.m²/C

The electric flux through any closed surface is proportional to the charge enclosed by the surface. The electric flux through an open surface can be calculated by multiplying the electric field intensity with the area of the surface and the cosine of the angle between the electric field intensity and the normal to the surface.

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Complete question:

A flat sheet of paper of area 0.130 m2 is oriented so that the normal to the sheet is at an angle of 56 ∘ to a uniform electric field of magnitude 18 N/C .Find the amount of elctric flux passing through the sheet?

The first radioiodine isotopes to be used to treat thyrotoxicosis were iodine-130 and iodine-131 in 1941, and thyroid cancer was added to their list of uses in 1943.

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The balanced equation for the combustion of ethane, C2H6, is: C2H6 + 3O2 → 2CO2 + 3H2OTo determine how much heat is produced if 7.0 moles of ethane undergo complete combustion, we need to use the balanced equation and the standard enthalpies of formation of the reactants and products.

The standard enthalpy of formation of a compound is the enthalpy change when one mole of the compound is formed from its constituent elements, with all reactants and products in their standard states (usually at 1 atm and 25°C).The standard enthalpies of formation of the reactants and products in the combustion of ethane are:

ΔHf°(C2H6) = -84.68 kJ/mol

ΔHf°(O2) = 0 kJ/mol

ΔHf°(CO2) = -393.51 kJ/mol

ΔHf°(H2O) = -285.83 kJ/mol

Now we can calculate the heat produced by using the difference between the enthalpies of the products and reactants:

2CO2 + 3H2O - (C2H6 + 3O2)

ΔH = 2(-393.51 kJ/mol) + 3(-285.83 kJ/mol) - (-84.68 kJ/mol + 3(0 kJ/mol))

ΔH = -1560.78 kJ/mol

Therefore, if 7.0 moles of ethane undergo complete combustion, the amount of heat produced will be:

-1560.78 kJ/mol x 7.0 mol

= -10,925.46 kJ or -10,925,460 J.

Note that the negative sign indicates that heat is released by the reaction, which is exothermic.

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Answer:

24 meters

Explanation:

24 meter because area

area=Lxw

so just multiply 4 x 6 = 24

Answer:

C.  12 meters2

Fuel shortages in the 1980s started the interest in fuel efficiency  best describes a society’s desire for fuel-efficient vehicles because of Iran- Iraq war.

The Iran-Iraq War in 1980 is the main cause of shortage of oil production. Iraq's and Iran's oil production dropped significantly which triggering economic recessions worldwide.

So that's why society’s desire for fuel-efficient vehicles in order to reduce consumption as well as reduce pollution.

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Answer:

0.00268 M

Explanation:

To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.

Step 1:

3.00 mL / 1,000 = 0.00300 L

Molarity = moles / volume (L)

0.0250 M = moles / 0.00300 L

(0.0250 M) x (0.00300 L) = moles

7.50 x 10⁻⁵ = moles

Step 2:

25.0 mL / 1,000 = 0.0250 L

0.0250 L + 0.00300 L = 0.0280 L

Molarity = moles / volume (L)

Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)

Molarity = 0.00268 M

Category ------------ Nuclear chemistry

Sub-category --------- Radioactivity

ANSWER

The remaining amount of isotope after 1600 years is 32.9444 grams

EXPLANATION

Given that;

The half life of the isotope is 5715 years

The initial mass of the isotope is 40 grams

The decay time is 1600 years

Follow the steps below to find the remaining mass after 1600 years

Where

A(t) is the amount of mass remaining after time t

Ao is the initial amount of mass of the substance

t1/2 is the half-life of the substance

t is the decay time

Substitute the given data into the above formula

Therefore, the remaining amount of isotope after 1600 years is 32.9444 grams

Answer:

1.4 L.

Explanation:

Applying Charles law,

V/T = V'/T'....................... Equation 1

Where V = Initial volume, V' = Final volume, T = Initial Temperature in Kelvin, T' = Final Temperature in Kelvin.

Make V' the subject of the equation

V' = (V/T)T'..................... Euqation 2

Given: V = 1.5 L, T = 295 K, T' = 2 °C = (2+273) K = 275 K

Substitute these values into equation 2

V' = (1.5/295)275

V' = 1.398

V' ≈ 1.4 L

The total charge that flows while the battery is being charged is approximately 193,939.39 Coulombs.

To calculate the total charge that flows while the battery is being charged, we can use the relationship between electrical energy, potential difference, and charge.

The electrical energy (E) stored in the battery is given as 64 mega Jules (64 MJ). The potential difference (V) provided by the battery is 330 volts. We know that the energy (E) is equal to the product of the potential difference (V) and the charge (Q):

E = V * Q

Since the charging process is 100% efficient, all the electrical energy supplied is stored in the battery. Therefore, we can rearrange the equation to solve for the charge (Q):

Q = E / V

Substituting the given values, we have:

Q = 64 MJ / 330 V

To perform the calculation, we need to convert mega Jules (MJ) to joules (J) since the SI unit of energy is joules. One mega Joule is equal to 1 million joules:

Q = (64 * 10^6 J) / 330 V

Calculating the division:

Q ≈ 193,939.39 Coulombs

Therefore, the total charge that flows while the battery is being charged is approximately 193,939.39 Coulombs.

This value represents the quantity of electric charge transferred during the charging process, and it indicates the amount of electricity that enters the battery.

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The isotope is a neutral atom as the number of electrons and protons are equal in an isotopic atom.

Isotopes can be described as atoms of an element that exhibit the same atomic number but a different atomic mass. Isotopes of an element exhibit an equal number of electrons and protons Therefore, they are not charged atoms. The numbers of neutrons in their respective nucleus of an isotope of an element is different.

In the isotopes of oxygen element, the number of neutrons in oxygen increases by 1, and it creates the isotope of oxygen. The number of electrons or protons in each isotope of oxygen element are eight therefore, all the isotope are neutral.

Therefore, the isotopes of an element are not charged atoms.

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To address the crevice corrosion issue in the high-pressure connectors and piping of the SWRO plant, several remedies can be considered, A SWRO (Sea Water Reverse Osmosis) plant is a water desalination facility that uses reverse osmosis technology to treat seawater or brackish water and produce freshwater

Material Selection: Evaluate the material compatibility with the operating conditions, especially the chloride ion concentration and temperature. Consider using corrosion-resistant alloys such as duplex stainless steel (e.g., 2205) or super duplex stainless steel (e.g., 2507) that have better resistance to chloride-induced corrosion compared to 316L or 317L stainless steel.

Surface Treatment: Apply appropriate surface treatments to enhance corrosion resistance. Passivation or pickling can remove surface contaminants and create a protective oxide layer on the metal surface, reducing the susceptibility to corrosion.

Design Modifications: Evaluate the design of the connectors and piping to minimize crevices and stagnant areas where corrosion can occur. Smooth transitions, avoiding sharp angles or crevices, can help promote better fluid flow and prevent the accumulation of corrosive substances.

Cathodic Protection: Implement cathodic protection methods, such as impressed current or sacrificial anodes, to protect the connectors and piping from corrosion. This technique involves introducing a more easily corroded metal (anode) to the system, which sacrifices itself to protect the connected metal (cathode) from corrosion.

Monitoring and Maintenance: Regularly monitor the corrosion levels and condition of the connectors and piping. Implement a maintenance program that includes periodic inspections, cleaning, and repairs, if necessary, to prevent corrosion from progressing.

It is important to consult with corrosion experts and engineers who specialize in SWRO plant operations to assess the specific conditions, perform material testing, and provide tailored solutions to mitigate the crevice corrosion problem effectively.

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Answer:

In the chemical industry.

Isotopes are the atoms of the same element with same atomic no: and different mass numbers. ... ⇒As isotope of uranium is used as a fuel in nuclear reactor. ⇒An isotope of iodine is used in the treatment of goitre. ⇒An isotope of cobalt is used in the treatment of cancer.

Answer:

C. Sodium Chloride.

Explanation:

Because when NaCl dissolved in water, it will dissociate to produce 2 ions Na+ and Cl-.

The concentration of dye in solution C was calculated to be 3.839 M.

Dilution is when the extra solvent is added to a solution without increasing the solute concentration. The dilution factor is an expression that describes the ratio of the aliquot to the final volume of the solution. The final solution should be well mixed to make sure that all components are the same.

The dilution factor is a factor used to dilute the stock solution.

Given the concentration of dye in solution A=21.729 M

Solution B-  Dilution factor = DF1 = final volume/aliquot volume

DF1 = 16/4 = 4

Solution C- Dilution factor = DF2 = 10/6 = 1.66

So the total dilution factor = DF1 + DF2 = 4 + 1.66 = 5.66

So the concentration of dye in solution C is calculated as

The concentration in solution A/ DF = 21.729/5.66= 3.839 M

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NaCl crystallizes in a cubic unit cell, meaning each side of the cell is equal in length. In this structure, each corner of the cell has a Cl- ion and each face of the cell has a Na+ ion, totaling 8 Cl- ions and 6 Na+ ions per unit cell. So, there are 8 Cl- ions and 6 Na+ ions in each unit cell of NaCl.

NaCl is an ionic compound, meaning it is composed of a metal cation (Na+) and a nonmetal anion (Cl-). Ionic compounds form when electrons are transferred from the metal to the nonmetal, creating a strong electrostatic attraction between the two ions. The oppositely charged ions are then arranged into a lattice structure with a repeating pattern of alternating cations and anions.

The unit cell of a NaCl crystal is a cube, meaning all the sides of the unit cell are of equal length. A single unit cell of NaCl contains eight Cl- ions at the corners and six Na+ ions at the faces of the cube. Since the unit cell of NaCl is symmetrical, the same arrangement of ions is repeated in each adjacent unit cell of the crystal. The arrangement of ions in a NaCl unit cell is important in understanding its properties. Because the cations and anions are arranged in an alternating pattern, the ions form a strong ionic bond. This bond gives the crystal its hardness and stability, making NaCl one of the most common and widely used compounds.

In summary, each unit cell of NaCl contains 8 Cl- ions at the corners and 6 Na+ ions at the faces, forming a cube of equal sides. The alternating arrangement of cations and anions creates an ionic bond which gives NaCl its hardness and stability.

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Answer:

livermorium

Explanation:

The atom with the largest number of neutrons is a tie between livermorium and tennessine. Each of these atoms contain 177 neutrons.

Answer:

162.7miles/hr

Explanation:

Given parameters:

Distance covered by the train  = 813.5miles

Time taken  = 5hours

Unknown:

Speed of the train  = ?

Solution:

Speed is a physical quantity.

It is mathematically expressed as;

      Speed  = \(\frac{distance}{time}\)

So, input parameters and solve;

   Speed  = \(\frac{813.5}{5}\)  = 162.7miles/hr